Tag: conic sections

Quizzes (12) Questions (54)

Questions Related to conic sections

Multiple choice mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

Find the locus of the point of intersection of the lines $\sqrt{3}x-y-4\sqrt{3} \lambda=0$ and $\sqrt{3}\lambda x+\lambda y-4\sqrt{3}=0$ for different values of $\lambda$.

  1. $4x^2-y^2=48$
  2. $x^2-4y^2=48$
  3. $3x^2-y^2=48$
  4. $y^2-3x^2=48$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let $(h,k)$ be the point of intersection of the given lines. Then,


$\sqrt{3}h-k-4\sqrt{3} \lambda=0$ and $\sqrt{3} \lambda h+\lambda k-4\sqrt{3}=0$


$\sqrt{3}h-k=4\sqrt{3}\lambda$ and $\lambda(\sqrt{3}h+k)=4\sqrt{3}$

$(\sqrt{3}h-k)\lambda(\sqrt{3}h+k)=(4\sqrt{3}\lambda)(4\sqrt{3})$

$3h^2-k^2=48$

Hence, the locus of (h,k) is $3x^2-y^2=48$.

Multiple choice mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

$Center\quad of\quad the\quad hyperbola\quad { x }^{ 2 }+4{ y }^{ 2 }+6xy+8x-2y+7=0\quad is\quad $

  1. $(1,1)$
  2. $(0,2)$
  3. $(2,0)$
  4. $None\quad of\quad these$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Given Hyperbola: $x^{2}+4y^{2}+6xy+8x-2y+7=0$
Centre: Point of intersection of asymptotes of hyperbola.
Now finding Asymptotes of given equation, taking equation of asymptote $y=mx+c$ by replacing $x\rightarrow 1, y\rightarrow m$ in $\phi _n(m)$
When $n=2$, $\phi _{2}(m)=1+4m^{2}+6m$
$\phi _{1}(m)=8-2m$
$\phi _{0}(m)=7$
$\phi _{2}^{1}(m)=8m+6$
$\phi _{1}(m)=8-2m$

Putting $\phi _{2}(m)=0$
$\Rightarrow 1+6m+4m^{2}=0$
$\Rightarrow  m=\cfrac{-6\pm \sqrt {36-16}}{2(4)}$
$\Rightarrow m=\cfrac{-6\pm\sqrt 20}{2(4)}$
$\Rightarrow m=\cfrac{-3\pm \sqrt 5}{4}$
So, m$=\cfrac{-3+\sqrt 5}{4}, \cfrac{-3-\sqrt 5}{4}$
Value of $c$, when $m$ is different
$c= \cfrac{-\phi _{1}(m)}{\phi _{2}^{'}(m)}=\cfrac{8-2m}{8m+6}$

For $m=\cfrac{-3+\sqrt 5}{4}, c=\cfrac{8-2(\cfrac{-3+\sqrt 5}{4})}{8\cfrac{-3+\sqrt 5}{4})+6}=\cfrac{3-\sqrt 5+16}{-6+2\sqrt 5+6}=\cfrac{19\sqrt 5-5}{10}$

For $m=\cfrac{-3-\sqrt 5}{4}, c=\cfrac{8-2(\cfrac{-3-\sqrt 5}{4})}{8\cfrac{-3-\sqrt 5}{4})+6}=\cfrac{19+15}{-2 \sqrt 5}=\cfrac{-(19\sqrt 5+5)}{10}$
Equation of Asymptotes : $y=\cfrac{-3+\sqrt 5}{4}x +\cfrac{19\sqrt 5-5}{10}$ & $y=\cfrac{-3-\sqrt 5}{4}x-(\cfrac{5+19\sqrt 5}{10})$
On solving them for x & y, putting LHS-RHS
$\Rightarrow 0=\cfrac{\sqrt 5}{2}x+\cfrac{19\sqrt 5}{5} \Rightarrow x=\cfrac{-38}{5}$
Now putting values of x in Asymptotes equation, we get
$y=\cfrac{-3-\sqrt 5(-19)}{10}+\cfrac{+5+19\sqrt 5}{10}=\cfrac{+52}{10}$
Centre$(\cfrac{-38}{5}, \cfrac{+52}{10})$.
Multiple choice mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

Circles are drawn on chords of the rectangular hyperbola $xy=4$ parallel to the line $y=x$ as diameters.All such circles pass through two fixed points whose coordinates are 

  1. $\left(2,2\right)$
  2. $\left(2,-2\right)$
  3. $\left(-2,2\right)$
  4. $\left(-2,-2\right)$
Reveal answer Fill a bubble to check yourself
A,D Correct answer
Explanation
Given:Rectangular hyperbola $xy=4={c}^{2}$
$\Rightarrow\,{c}^{2}=4$
$\Rightarrow\,c=2$
Let $P$ and $Q$ be the end points on the Rectangular hyperbola where $P\left(2{t} _{1},\dfrac{2}{{t} _{1}}\right)$ and $Q\left(2{t} _{2},\dfrac{2}{{t} _{2}}\right)$
Using the end points of diameter the equation of the circle

$C:\left(x-2{t} _{1}\right)\left(x-2{t} _{2}\right)+\left(y-\dfrac{2}{{t} _{1}}\right)\left(y-\dfrac{2}{{t} _{2}}\right)=1$         ..........$(1)$

Now,Slope of the $PQ=\dfrac{\dfrac{2}{{t} _{2}}-\dfrac{2}{{t} _{1}}}{2{t} _{2}-2{t} _{1}}$

$=\dfrac{2\left(\dfrac{1}{{t} _{2}}-\dfrac{1}{{t} _{1}}\right)}{2\left({t} _{2}-{t} _{1}\right)}$

$=\dfrac{\dfrac{{t} _{1}-{t} _{2}}{{t} _{1}{t} _{2}}}{\left({t} _{2}-{t} _{1}\right)}$

$=\dfrac{\dfrac{{t} _{1}-{t} _{2}}{{t} _{1}{t} _{2}}}{\left({t} _{2}-{t} _{1}\right)}$

$=\dfrac{-1}{{t} _{1}{t} _{2}}$

Hence $PQ$ is the diameter for circle and it is parallel to the line $y=x$

Slope of $PQ=$Slope of the line $y=x$

$\Rightarrow\,\dfrac{-1}{{t} _{1}{t} _{2}}=1$

$\Rightarrow\,{t} _{1}{t} _{2}=-1$

$(1)\Rightarrow\,\left(x-2{t} _{1}\right)\left(x-2{t} _{2}\right)+\left(y-\dfrac{2}{{t} _{1}}\right)\left(y-\dfrac{2}{{t} _{2}}\right)=1$ 

$\Rightarrow\,x\left(x-2{t} _{2}\right)-2{t} _{1}\left(x-2{t} _{2}\right)+y\left(y-\dfrac{2}{{t} _{2}}\right)-\dfrac{2}{{t} _{1}}\left(y-\dfrac{2}{{t} _{2}}\right)=1$
 
$\Rightarrow\,{x}^{2}-2x{t} _{2}-2x{t} _{1}+4{t} _{1}{t} _{2}+{y}^{2}-\dfrac{2y}{{t} _{2}}-\dfrac{2y}{{t} _{1}}+\dfrac{4}{{t} _{1}{t} _{2}}=1$

$\Rightarrow\,{x}^{2}-2x{t} _{2}-2x{t} _{1}+4\times -1+{y}^{2}-\dfrac{2y}{{t} _{2}}-\dfrac{2y}{{t} _{1}}+\dfrac{4}{\times -1}=1$ using ${t} _{1}{t} _{2}=-1$

$\Rightarrow\,{x}^{2}+{y}^{2}-2x\left({t} _{2}+{t} _{1}\right)-4-2y\left(\dfrac{1}{{t} _{2}}+\dfrac{1}{{t} _{1}}\right)-4=1$

$\Rightarrow\,{x}^{2}+{y}^{2}-8-2x\left({t} _{2}+{t} _{1}\right)-2y\left(\dfrac{{t} _{1}+{t} _{2}}{{t} _{1}{t} _{2}}\right)=1$

$\Rightarrow\,{x}^{2}+{y}^{2}-8-2x\left({t} _{2}+{t} _{1}\right)-2y\left(\dfrac{{t} _{1}+{t} _{2}}{-1}\right)=1$ using ${t} _{1}{t} _{2}=-1$

$\Rightarrow\,{x}^{2}+{y}^{2}-8-2x\left({t} _{2}+{t} _{1}\right)+2y\left({t} _{1}+{t} _{2}\right)=1$ 

$\Rightarrow\,{x}^{2}+{y}^{2}-8+\left(2y-2x\right)\left({t} _{2}+{t} _{1}\right)=1$ is of the form $C+\lambda\,L$ 

where $C={x}^{2}+{y}^{2}-8=0$ is the equation of a circle.
and $L=2y-2x=0$ is the equation of a line.
$\Rightarrow\,y-x=0$ or $x=y$

Substituting $x=y$ in the equation ${x}^{2}+{y}^{2}-8=0$ we get
$\Rightarrow\,2{x}^{2}-8=0$

$\Rightarrow\,2\left({x}^{2}-4\right)=0$

$\Rightarrow\,\left(x-2\right)\left(x+2\right)=0$

$\therefore\,x=2,-2$

$\Rightarrow\,y=2,-2$ since $x=y$

Hence the coordinates of the fixed points are $\left(2,2\right)$ and $\left(-2,-2\right)$ 
Multiple choice mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

Centre of the hyperbola ${x^2} + 4{y^2} + 6xy + 8x - 2y + 7 = 0$ is 

  1. $(1,1)$
  2. $(0,2)$
  3. $(2,0)$

  4. None of these
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Consider equation of a hyperbola as $F=ax^2+2by+cy^2+2dx+2ey+f=0$

The centre of this hyperbola can be found by applying the concepts of partial differentiation
We first find $\dfrac{\delta F}{\delta x} $ and $\dfrac{\delta F}{\delta y}$
We then solve $\dfrac{\delta F}{\delta x} =0$ and $\dfrac{\delta F}{\delta y}=0 $ to find $x,y$ which is the centre of the hyperbola .

Given that,
$F=x^2+4y^2+6xy+8x-2y+7$

$\Rightarrow \dfrac{\delta F}{\delta x}=2x+0+6y+8+0+0$

$\Rightarrow \dfrac{\delta F}{\delta x}=2x+6y+8$         ...$(1)$

$\Rightarrow \dfrac{\delta F}{\delta y}=0+8y+6x+0-2+0$

$\Rightarrow \dfrac{\delta F}{\delta y}=6x+8y-2$      ....$(2)$
 
$(1) \rightarrow 2x+6y+8=0$                    

$(2) \rightarrow 6x+8y-2=0$

Solving $(1), (2)$ we get,

$\Rightarrow x=\dfrac{19}{5}, y=\dfrac{-13}{5}$

Therefore the centre of hyperbola is $(\dfrac{19}{5},\dfrac{-13}{5})$



Multiple choice mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

The eccentricity of the hyperbola whose latus-return is $8$ and length of the conjugate axis is equal to half the distance between the foci, is

  1. $\dfrac43$
  2. $\dfrac4{\surd 3}$
  3. $\dfrac2{\surd 3}$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given that the length of the latus rectum is $8$ and length of the conjugate axis is equal to half the distance between the foci.


$\Rightarrow \dfrac{2b^2}{a}=8$ and $2b=\dfrac{1}{2}(2ae)$

$\therefore \dfrac{2}{a}\left(\dfrac{ae}{2}\right)^2=8$

$\Rightarrow ae^2=16$ ...(1)

We have $\dfrac{2b^2}{a}=8$

$\Rightarrow b^2=4a$

$\Rightarrow a^2(e^2-1)=4a$

$\Rightarrow ae^2-a=4$

$\Rightarrow 16-a=4$ (by (1))

$\Rightarrow a=12$

Substitute $a=12$ in (1)

$\Rightarrow 12e^2=16$

$\Rightarrow e^2=\dfrac{4}{3}$

$\therefore e=\dfrac{2}{\sqrt{3}}$

Multiple choice mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

From any point on the hyperbola $\displaystyle \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ tangents are drawn to the hyperbola $\displaystyle \frac{x^2}{a^2} - \frac{y^2}{b^2} = 2$. The area cut-off by the chord of contact on the asymptotes is equal to

  1. $\displaystyle \frac{ab}{2}$

  2. ab

  3. 2 ab

  4. 4 ab
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let $P\left( { x } _{ 1 },{ y } _{ 1 } \right) $ be a point on the hyperbola $\cfrac { { x }^{ 2 } }{ { a }^{ 2 } } -\cfrac { { y }^{ 2 }
}{ { b }^{ 2 } } =1$. Then,
$\cfrac { { { x } _{ 1 } }^{ 2 } }{ { a }^{ 2 } } -\cfrac { { { y } _{ 1 } }^{ 2 } }{ { b }^{ 2 } } =1$
The chord of contact of tangents from $P$ to the hyperbola $\cfrac { { x

}^{ 2 } }{ { a }^{ 2 } } -\cfrac { { y }^{ 2 } }{ { b }^{ 2 } } =2$ is
$\cfrac

{ { x }{ x } _{ 1 } }{ { a }^{ 2 } } -\cfrac { { y }{ y } _{ 1 } }{ { b

}^{ 2 } } =2\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot (i)$

The equations of the asymptotes are
$\cfrac { x }{ a } -\cfrac { y }{ b } =0$ and $\cfrac { x }{ a } +\cfrac { y }{ b } =0$

The points of intersection of $(i)$ with the two asymptotes are given by
${

x } _{ 1 }=\cfrac { 2a }{ \cfrac { { x } _{ 1 } }{ a } -\cfrac { { \quad y

} _{ 1 } }{ b }  } \quad ,{ \quad y } _{ 1 }=\cfrac { 2b }{ \cfrac { { x

} _{ 1 } }{ a } -\cfrac { { \quad y } _{ 1 } }{ b }  } $
${ x } _{ 2

}=\cfrac { 2a }{ \cfrac { { x } _{ 1 } }{ a } -\cfrac { { \quad y } _{ 1 }

}{ b }  } \quad ,{ \quad y } _{ 2 }=\cfrac {- 2b }{ \cfrac { { x } _{ 1 }

}{ a } -\cfrac { { \quad y } _{ 1 } }{ b }  } $
$\quad \therefore $ Area of the triangle
$\cfrac

{ 1 }{ 2 } \left( { x } _{ 1 }{ y } _{ 2 }-{ x } _{ 2 }{ y } _{ 1 } \right)

=\cfrac { 1 }{ 2 } \left( \cfrac { 4ab\times 2 }{ \cfrac { { { x } _{ 1 }

}^{ 2 } }{ { a }^{ 2 } } +\cfrac { { { y } _{ 1 } }^{ 2 } }{ { b }^{ 2 }

}  }  \right) =4ab$

Multiple choice mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

Let $a, b$ be non-zero real numbers. The equation $\displaystyle \left ( ax^{2}+by^{2}+c \right )\left ( x^{2}-5xy+6y^{2} \right )$ represents

  1. four straight lines, when $\displaystyle c=0$ and $a, b$ are of the same sign
  2. two straight lines and a circle, when $\displaystyle a=b$ and $c$ is of sign opposite to that of $a$
  3. two straight lines and a hyperbola, when $a$ and $b$ are of the same sign and $c$ is of sign opposite to that of $a$
  4. a circle and an ellipse, when $a$ and $b$ are of the same sign
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

From given expression, $\displaystyle { x }^{ 2 }-5xy+6{ y }^{ 2 }=0$ are pair of straight lines $\displaystyle y=\frac { x }{ 2 } $ and $ y=\dfrac { x }{ 3 } $
Now, $\displaystyle a{ x }^{ 2 }+b{ y }^{ 2 }+c=0$ will be cirlce of radius $\displaystyle \sqrt { -\frac { c }{ a }  } $.

If $\displaystyle a=b$ and $\displaystyle c$ is of opposite sign of $\displaystyle a  $ and $  b$, then $\displaystyle { x }^{ 2 }+{ y }^{ 2 }=-\frac { c }{ a } $

Multiple choice mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

If a hyperbola passes through the foci of the ellipse $\displaystyle \frac {x^2}{25} + \frac {y^2}{16} = 1$ and its traverse and conjugate axis coincide with major and minor axes of the ellipse, and product of the eccentricities is 1, then:

  1. Equations of the hyperbola is $\displaystyle \frac {x^2}{9} - \frac {y^2}{16} = 1$
  2. Equations of the hyperbola is $\displaystyle \frac {x^2}{9} - \frac {y^2}{25} = 1$
  3. Focus of the hyperbola is $\displaystyle (5, 0)$
  4. Focus of the hyperbola is $\displaystyle (5 \sqrt 3, 0)$
Reveal answer Fill a bubble to check yourself
A,C Correct answer
Explanation

Given ellipse is, $\displaystyle \frac {x^2}{25} + \frac {y^2}{16} = 1$
Eccentricity of the ellipse is, $\displaystyle e _e  =\sqrt{1-\frac{b^2}{a^2}}=\frac{3}{5}$
So the foci of the ellipse is, $(\pm ae _e,0)=(\pm 3 , 0)$
Let eccentricity of the required  hyperbola is $e _h$ and semi major and minor axes are $a$ and $b$, so the equation of hyperbola is, $\displaystyle \frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
Given hyperbola passes trough $(\pm 3,0)\Rightarrow \displaystyle \frac{9}{a^2}-\frac{0}{b^2}=1\Rightarrow a^2 = 9$
Also given that $\displaystyle  e _e\times e _h = 1$ $\Rightarrow e _h=\cfrac{5}{3}$ $\Rightarrow$ $b^2 =a^2(e _h^2-1)=16 $
Hence required hyperbola is, $\displaystyle \frac{x^2}{9}-\frac{y^2}{16}=1$
And foci of the hyperbola is, $(\pm 5,0)$

Multiple choice mathematics and statistics hyperbola introduction to hyperbola standard equation of hyperbola conic sections

The equation ${x}^{2}+9=2{y}^{2}$ is an example of which of the following curves?

  1. hyperbola

  2. circle

  3. ellipse

  4. parabola

  5. line
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given, ${x}^{2}+9=2{y}^{2}$ 

$\Rightarrow 2{y}^{2}-{x}^{2}=9$
$\Rightarrow \dfrac { { y }^{ 2 } }{ 9/2 } -\dfrac { { x }^{ 2 } }{ 9 } =1$
It is in the form of $\dfrac { { y }^{ 2 } }{ { a }^{ 2 } } -\dfrac { { x }^{ 2 } }{ { b }^{ 2 } } =1$ which is the equation of hyperbola.
Therefore, the given equation is a equation of hyperbola.